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Math 131A  Section 2
Spring 2010
Homework 7
The following problems were graded on this assignment: 18.4 (2 points), 18.8 (2 points), 18.12 (3 points), 19.1(b,e) (2
points), 19.4 (3 points), 19.7 (3 points)
18.4. Let
f
(
x
) = (
x

x
0
)

1
. Observe that this function is continuous for every
x
6
=
x
0
by 17.4, so
f
is continuous on
S
since
x
0
/
∈
S
. Let
M >
0 and let
±
=
M

1
. Then, since lim
x
n
=
x
0
, there is
N
such that

x
n

x
0

< ±
for
n > N
.
Then,

f
(
x
N
+1
)

=

x
N
+1

x
0


1
> M
, so
f
is unbounded on
S
.
18.8. If
a
=
b
, then
f
(
a
)
f
(
b
) =
f
(
a
)
2
≥
0, a contradiction. Hence, we may assume without loss of generality that
a < b
.
Note that
f
(
a
)
6
= 0, since this implies
f
(
a
)
f
(
b
) = 0. Suppose
f
(
a
)
<
0. Then, since
f
(
a
)
f
(
b
)
<
0, we have that
f
(
b
)
>
0.
Then, 0
∈
(
f
(
a
)
,f
(
b
)), so there exists
x
∈
(
a,b
) such that
f
(
x
) = 0 by the intermediate value theorem. Similarly, if
f
(
a
)
>
0, then
f
(
b
)
<
0, so there exists
x
∈
(
a,b
) such that
f
(
x
) = 0 since 0
∈
(
f
(
b
)
,f
(
a
)) by the intermediate value
theorem, giving the desired result.
18.12.(a). This result is shown in the last homework assignment, so a solution can be seen there.
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This note was uploaded on 11/20/2010 for the course MATH 131A 131A taught by Professor Kim during the Spring '10 term at UCLA.
 Spring '10
 Kim

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